A pair of numbers is picked up randomly (without replacement) from the set {1, 2, 3,5, 7, 11, 12, 13, 17, 19). The probability that the number 11 was picked given that the sum of the numbers was even is nearly

Let Abe the event that 11 is picked and B be the event that sum is even. The number of ways of selecting 11 along with one more odd number is $\mathrm{n}(\mathrm{A}\cap \mathrm{B}){=}^7{\mathrm{C}}_1$. 

The number of ways of selecting either two even numbers or selecting two odd numbers is $\mathrm{n}\left(\mathrm{B}\right)=1{+}^8{\mathrm{C}}_2$. 

$\therefore \mathrm{P}(\mathrm{A}/\mathrm{B})=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}=\frac{7}{29}=0.24$